# I do not understand how the Work done by gravity on a freely falling body is equal to the negative of the change in potential energy. Pls read below

I have tried to understand how the change in potential energy is equal to the negative of the work done by gravity on a body in free fall.

If we were to consider a body of mass $$m$$ dropped from height $$h_1$$ to $$h_2$$ and try to use $$E_g = -(U_f - U_i)$$ where $$W_g$$ is the work done by gravity, $$U_f$$ is the final potential energy and $$U_i$$ is the initial potential energy, then:

$$W_g=mg(h_2-h_1)$$

$$U_f-U_i=mgh_2-mgh_1=mg(h_2-h_1)$$

In which Work done by gravity is clearly NOT EQUAL to the negative of the change in potential energy. Am I doing something wrong here?

However, if were to to consider the opposite motion of the body being lifed by us from height $$h_2$$ to $$h_1$$, $$W_u$$ is the work done by us, $$U_f$$ is the final potential energy and $$U_i$$ is the initial potential energy, then:

$$Wu=-(mg(h_1-h_2))$$ (We add negative sigh here since displacement is in the opposite direction of force applied by us.)

$$U_f-U_i=mgh_1-mgh_2=mg(h_1-h_2)$$

Here the statement 'Work done by gravity is the negative of the change in potential energy' holds true, but not in the first case. Please could you explain this.

• Remember that all we mean by the term 'potential energy' is that something has the capacity to do work. So when we say "this mass has a gravitational potential energy of mgh" what we are saying is that if we let the object descend by a distance h, gravity will do work of $mgh$ Joules. Jan 11 at 20:08
• The change in potential energy is defined to be minus the work done by the force, so if you ever do a calculation where these don't agree with each other, you are doing something wrong.
– d_b
Jan 11 at 21:02
• @Chemomechanics: That looks like an answer to me. Jan 11 at 21:06

## 3 Answers

You are considering:

a body of mass $$m$$ dropped from height $$h_1$$ to $$h_2$$

For the sake of illustration, let's say it falls from $$h_1=10$$m to $$h_2=5$$m. The distance moved is therefore $$5$$m downwards ($$\Delta h=-5$$m).

You write:

$$W_g=mg(h_2-h_1)$$

But the acceleration due to gravity '$$g$$' is also downwards; so you need to use a value of something like $$-9.8ms^{-2}$$. This makes $$W_g$$ come out as positive.

• @SuhasBharadwaj I have edited my answer to clarify the precise point where your reasoning needs to change - is there anything else that is confusing you? Jan 12 at 12:06
• Thank you so much sir. However if we take 'g' as negative then in the case that the object moves up, then the work done by gravity would be Wg=-[-mg(h2-h1)] (Here, I put a negative sign out of the square brackets beacuse work done is negative and as per your suggestion in your answer, I have put the acceleration due to gravity as -g). Thus Wg=mg(h2-h1) and therefore the statement would hold untrue for the case of the body moving up. Maybe I shouldn't add the extra negative sign outside the square bracket. But I don't understand why we shouldn't indicate negative work done. Jan 12 at 20:26

I have tried to understand how the change in potential energy is equal to the negative of the work done by gravity on a body in free fall.

If the gravitational force does positive work then the gravitational potential energy (a measure of the capacity to do work) of the system will decrease.

Let $$\hat u$$ be the unit vector pointing upwards (but please note you will get the same result if the unit vector down, $$\hat d$$, had been chosen).

The work done by the gravitational field is

$$m \vec g \cdot \Delta \vec h = - mg \hat u \cdot (h_{\rm final} \hat u - h_{\rm initial}\hat u) = -mg (h_{\rm final} - h_{\rm initial})$$

and so the change in gravitational potential energy is given by

$$\Delta U = -(-mg (h_{\rm final} - h_{\rm initial}))=mg (h_{\rm final} - h_{\rm initial})$$

If the body falls then $$h_{\rm final} and $$\Delta U$$ is negative, ie the gravitational potential energy decreases.

If the body rises then $$h_{\rm final}>h_{\rm initial}$$ and $$\Delta U$$ is positive, ie the gravitational potential energy increases.

Here the statement 'Work done by gravity is the negative of the change in potential energy' holds true, but not in the first case. Please could you explain this.

There are two issues that I see. The first is a math error between the first equation and second equation. The second is not recognizing that in your second example it is the negative work done by gravity when a body is raised that is responsible for the gravitational potential energy. To elaborate:

FREE FALLING BODY:

In the first equation I'm assuming $$E_{g}$$ is supposed to be $$W_{g}$$. In any case, since $$U_{f}=mgh_{2}$$ and $$U_{i}=mgh_{1}$$ your second equation is inconsistent with your first equation. Your second equation should read

$$W_g=mg(h_1-h_2)\tag{1}$$

Note that the work done by gravity on the free falling body is positive work since the direction of the force of gravity is the same as the displacement of the body.

LIFTING THE BODY:

The work you do lifting the body is positive work since the direction of your force is the same as the direction of the displacement of the body, or,

$$W_{u}=mg(h_{1}-h_{2})$$

At the same time you are doing positive work, gravity does an equal amount of negative work (assuming the body is brought to rest at $$h_1$$.) Its work is negative because the direction of the force of gravity is opposite to the displacement of the body. So the work done by gravity is

$$W_{g}=-mg(h_{1}-h_{2})=mg(h_{2}-h_{1})\tag{2}$$

When something does negative work on a body it takes energy away from the body. In this case, gravity takes the energy you gave the body lifting it and stores it as gravitational potential energy.

Note that equations (1) and (2) are identical except for the sign. And that's because the work done by gravity is positive in equation (1) (the free falling body) and negative in equation (2) (the body being raised).

Hope this helps.